 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
1
Stress Transformation, Principal Stresses and
Mohr’s Circle (Chapter 5)
Consider Conservation of Linear Momentum and assume
there is no mass flux and that body forces are negligible
(only the traction terms remain). The result is static
equilibrium of stresses on a differential volume:
 xx  yx  zx


0
x
y
z
 xy  yy  zy


0
x
y
z
 xz  yz  zz


0
x
y
z
 2001, W. E. Haisler
2
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
Consider a stress state where the only non-zero stresses applied
to the volume occur in a 2-D plane oriented along the coordinate
axes. This is called plane stress. For example, we can have
plane stress in the x-y plane, the x-z plane, or the y-z plane.
For example, plane stress in the x-y plane is shown below:
 xx
y
 yy
 zz
 yx
 xy 
 xy
 zz
xx
 xx
 xy
 yx
 yy
x generalized
z
 yy
 yx
plane stress
 yx

yy
plane stress
 xy
 xx
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
3
Assume plane stress (stresses in x-y plane only) so that the
traction (stress) tensor becomes:
 xx  xy



   yx  yy
0
 0
0
0

0 
We now consider the question of resolving the given stress
components in the x-y directions into stresses oriented in a
different direction. For example, the resultant stresses on a plane
which has a normal which makes an angle  with the x-axis. A
further question is whether there is some plane where the stresses
are a maximum or are zero.
Consider the following example:
 2001, W. E. Haisler
4
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
Consider a column loaded by a compressive pressure (traction) as
shown below in left figure. Draw two different free-bodys as shown:
P
y
x
column with compressive
load
P
 yy
freebody 1
P
 ns
 nn
 yy
 ns
 nn
freebody 2
For free-body 1, where we cut the structure normal to the y axis, we
obtain only the normal stress  yy on the cutting plane. However, in
fb#2, both normal  nn and shear  ns stresses exist.
 2001, W. E. Haisler
5
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
In the previous column example, we see that for a column loaded
only in compression, if one takes a cutting plane at some angle,
then shear stresses must exist on the cut plane. Consider a more
general problem.
F2
F1
F
F1
1
p
p
p
point 0
F2
F2
y
y’
x
x’

Suppose that we want to determine the internal stresses [ ] at
some point “O” in the body. We could use an x-y coordinate
system as shown in the middle figure, but we could also use and
x’-y’ coordinate system as shown in the right figure. Will we get
different results for [ ]. DEFINITELY, YES!
 2001, W. E. Haisler
6
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
Consider a solid body such as that shown below. Suppose that
we start with the state of stress defined in x-y coordinates.
F2
F1
 yy
 y' y'
p
 x' y'
 xy
point 0
y
x
z
 x'x'
 xx
 xx
 xy
 x' x'
 yy
 x' y'
y
x'
y'

x
 y' y'
 2001, W. E. Haisler
7
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
We wish to determine the state of stress in the x'-y' coordinate
system. We pass a cutting plane through point "O" which a unit
normal vector n as shown below to obtain:
 yy  zz
 yx
 xy
 xx
 xy
 zz
y’
 xx
 xx
 yx
 yy
n’
 xy
x'

x
z
n

y
y'
t (n )
 yx
 yy
x’
 2001, W. E. Haisler
8
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
A 2-D picture of the stress-state may be easier to work with:
 yy
 yx
 xx
 xy
y
x
z
 yx
 yy
 xy
 xx
 xx 
 xy
 yx
t( n )
n

y’ y
x’

x
 yy
Note that the unit vector n as well as the x'-axis makes an angle 
with the x-axis (measured CCW from the x-axis). From our
work on tractions in chapter 4, we have:
 2001, W. E. Haisler
9
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
y
n  nxi  n y j  cos i  sin j
t (n)
 xx
n


x
t( n)  t( n) i  t( n) j = traction
x
y
vector
 xy
 yy
 xx , yy , xy = Cauchy stresses
From Cauchy's formula, we have t( n )  n   or t(n)    n  


t
  xx cos   yx sin  n  xx  n  yx
( n)
x
y
x
t
  xy cos   yy sin  n  xy  n  yy
( n) y
x
y
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
The expression t
( n)
10
 n   is a general result in 3-D which
gives the projection of Cauchy stress tensor  onto a plane
whose unit normal is given by n .
The traction vector t( n) on the inclined face, as given above, is
written terms of it's x and y components. It is much more
informative and useful to write the traction vector t( n) in terms a
normal component  n and a parallel (shear) component  s as
shown below. Or, because the unit normal n and x'-axis are in
the same direction, we would actually be determining the stresses
 x ' x ' and  x ' y ' in the x'-y' coordinate system as shown below:
 2001, W. E. Haisler
11
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
t (n ) y
t( n )
n

 xx 
 xy
 yx
t (n ) x
 yy
 x' y'   s
 x' y'
t( n )

 x'x'   n
y’ y

x’
 = CCW angle from x-axis
The normal component  x ' x ' (also called  n ) is first obtained
from the dot product of the unit normal and the traction vector:
 n  n  tn
or as a vector  n  (n  tn )n
x
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
12
and the shear (parallel) component is obtained from vector
addition
tn   n   s

 s  t 
n
n
Now, lets carry out these vector operations to obtain  n and
 s . The unit normal vector in 2-D (x-y) is given by
n  nxi  n y j  cos i  sin j
Cauchy's formula in vector notation
tn  n    (nx xx  n y yx )i  (nx xy  n y yy ) j
 ( xx cos   xy sin  )i  ( xy cos   yy sin  ) j
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
13
Normal component of traction
 ( xx cos   xy sin  )i 
 x ' x '  tn  n  
 cos i  sin  j 
 ( xy cos   yy sin  ) j  


  cos2   2 sin  cos   sin 2 
xx
xy
yy
Use the double angle trig identities to rewrite above equation
cos 2   (1  cos 2 ) / 2
sin 2   (1  cos 2 ) / 2
2sin  cos  sin 2
 2001, W. E. Haisler
14
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
With the double angle trig identities,  x ' x ' becomes
 x'x'  n 
 xx   yy  xx   yy
2

2
cos 2   xy sin 2
Shear component of traction
The shear component,  s , can be obtained from vector
algebra, ie, tn   n 2   s
2
or
 s  tn   n2
2
or  x ' y '  tn   x2' x '
Evaluating the above and using the double angle formulas,
we obtain for the shear component:
2
 x ' y '   s   xy cos 2 
 xx   yy
2
sin 2
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
15
An alternate procedure to obtain the shear component:
Define the unit normal in the direction of y' to be n ' :
n '  k  n   sin i  cos j
The shear component  x ' y ' is the component of tn in the
direction of n ' so that
 ( xx cos   xy sin  )i 
 x ' y '  tn  n '  
   sin  i  cos j 

  ( xy cos   yy sin  ) j 


  xx cos sin    xy ( sin 2   cos 2  )   yy sin  cos
  xy cos 2 
 xx   yy
2
sin 2
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
16
These last two results allow us to transform the stresses from
an x-y coordinate system to an x'-y' coordinate system and
are called the stress transformation equations.
We could obviously use either the notation  x ' x ' or  n , and
 x ' y ' or  S . We will choose to use  n and  S . After squaring
both sides of the  n and  s equations, adding the results to
obtain one equation and using trig identities, we obtain the
following
2
2







xx
yy 
yy 
2
2  xx
  s   xy  
 n 


2
2




The above is similar to the equation of a circle of radius r
located at x=a and y=b, ie,
(x-a)2 + (y-b)2 = r2
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
17
Thus, the relation defining the normal and shear components
(in terms of the x and y stresses) can be drawn as a circle if
we choose the following:
x  n
a
 xx   yy
y  s
2
b0
2



 xx
yy 
r   xy  

2
2


This leads to graphical representation of the stress
transformation equations known as Mohr's Circle.
 2001, W. E. Haisler
18
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
 x ' y ' ( s )
  xx   yy 
2
r 



xy
2


2
 yy , xy 
 C

y  face stresses 
 P2
 P1
A
r
2 S
 Smax
a  ( xx   yy ) / 2
2 P

 x ' x ' ( n )

  xx ,  xy

B  x  face stresses

 P1 ,  P2  principal stresses
 is positive counter clockwise
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
19
Important note on sign convention for shear stress used
in constructing Mohr’s Circle.
When defining the Cauchy stress, a positive shear stress  xy on
the positive x face was in the positive y coordinate direction.
Because the calculation for the shear component above, ie,

s
 t
2
n
  2 which involves a square root, there is an
n
uncertainty in the + direction of the shear component. In order
for the Mohr’s circle graphical representation to be used
properly, we must adopt a sign convention:
Shear stresses on opposite faces that form a clockwise couple
about the center are positive on the Mohr’s circle.
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
 yy
 yx
 xy
 xx
 yy
 yx
 xy
 xx
20
Shear which makes
CW moment is positive
Differential volume + direction
Mohr's Circle
Shear on +x face is negative
Shear on +y face is positive
A positive  xy (in the stress tensor) is plotted on Mohr's
circle as negative for the x-face and positive for y-face.
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
21
Mohr’s Circle graphically represents the Cauchy formula for
tn  n   . It allows one to graphically determine the normal
and shear stress on any plane relative to the x-y axes.
Its most important use is to determine the principal stresses
(the maximum and minimum values of the normal stress,
 n , where there is no shear stress), the maximum shear
stress,  s , and the orientation of the planes on which these
occur.
The principal stress is the normal stress that occurs on a
plane where no shear stress exists.
 2001, W. E. Haisler
22
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
Construction of Mohr’s Circle (method 1)
1. Locate center on  n axis at ( xx   yy ) / 2.
2



 xx
yy 
2. Draw circle with radius r   xy  

2
2
 .

3. Locate the two points on circle with values of the stress
components on the x-face and y-face. These two points lie
on a diameter line of circle which passes through the center.
4. Determine max/min values of  n and  s and their plane
orientation (angle from x or y face). Remember: angles on
Mohr’s circle are twice the real world.
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
23
Construction of Mohr’s Circle (method 2)
1. Plot the values of normal and shear stress from the x-face ( xx and
 xy ) on the  n and  s axes. Observe Mohr’s circle assumption
on positive shear (shear is positive if moment due to shear is CW).
2. Plot the values of normal and shear stress from the y-face ( yy and
 yx ) on the  n and  s axes.
3. The above two points form the diameter of Mohr’s circle whose
center is located where the diameter line intersects the  n axis.
4. Determine max/min values of  n and  s and their plane
orientation (angle from x or y face). Remember: angles on Mohr’s
circle are twice the real world.
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
24
Some reminders:
1. For determining  n and  s from the defining equations, the
angle  between the normal to any plane and the x-axis is
defined positive in the CCW direction from the x-axis.
2. When plotting shear stresses on Mohr’s circle, a shear
stress is considered positive if it produces a CW moment.
3. All angles on Mohr’s circle are twice the real world [the
x- and y-face stresses are 90 in the x-y coordinate system,
but 180 apart on Mohr’s circle (opposite ends of circle
diameter)].
4. The planes of principal stress and maximum shear stress
are 90 apart on Mohr’s circle and thus 45 apart in the real
world!!!!
5. The angle  defining the planes where the maximum values of
 n and  s occur can also be obtained by calculus. If
 2001, W. E. Haisler
n 
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
 xx   yy  xx   yy
2

2
25
cos 2   xy sin 2
Then taking the derivative with respect  and setting equal to
zero gives the max/min values:
d n
 0  ( xx   yy )sin 2  2 xy cos 2
d
The above can be solved for two roots in the range: -90+90
which define the normal for the planes of max/min normal stress
with respect to the x-axis. (see the MAPLE example)
The max value of shear stress can be obtained in a similar
manner. The  defining plane of max shear is always 45 from
plane of max/min normal stress.
 2001, W. E. Haisler
26
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
Mohr's Circle Example. Consider the plane stress state given by
the stress tensor:
20 ksi
10 ksi
50 10 
ksi
  

10 20
50 ksi
50 ksi
10 ksi
20 ksi
1. Determine center of circle, point A.
A 
 xx   yy
2

50  20
 35ksi
2
2. Determine point B, x-face  ( xx ,  xy ) =(50,-10)
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
3. Determine point C, y-face  ( yy , xy ) =(20,+10)
4. Draw Mohr’s circle (plot x face first with  x ' y '  10 :
 x' y ' ( s )
50 10 
ksi
  

10 20 
(20,10)
y - face
 x' x' ( n )
(35, 0)
(50, -10)
x - face
27
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
28
5. Compute principal stresses and max shear stress.
  xx   yy 
 50  20 
2
2
r 




10
 18.03ksi


xy

2
 2 


2
2
 P   A  r  35  18.03ksi
 P1  35  18.03  53.03ksi
 P 2  35  18.03  16.97ksi
 Smax   r  18.03ksi
Note that we have two principal stresses:  P1 and  P 2 . These
are located 180 on Mohr's circle, but 90 in the real world. In
relation to the stress transformation equations,  P1 is the normal
stress in the x'-axis direction (which is oriented at an angle  P to
the x-axis) and  P 2 is the normal stress in the y'-axis direction.
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
29
The planes of maximum shear occur 45 (real world) from the
planes of principal stress.
Plot the principal stresses and max shear stresses on Mohr’s Circle:
 x' y ' ( s )

 18.03ksi
(35,18.03)
50 10 
ksi
  

10 20 
(35, 0)
 x' x' ( n )
Smax
(20,10)
y - face

P2
 16.97ksi
2 P
2 S

P1
(50, -10)
x - face
(35, 18.03)

Smax
 18.03ksi
 53.03ksi
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
30
6. Identify orientation of planes for principal stresses relative to
x-face plane.
 xy 
10 
1 
1 
2 P  tan 
 tan 
  33.7 deg

 50  35 
  xx   A 
 P  16.85deg CCW from x-face
The plane for the second principal stress  P 2 is ( P  90) deg
CCW from x-face (real world). Note: In the above calculation, a
formula was written down and used for 2 P . This is discouraged.
It is far simpler (and usually less mistakes) to look at Mohr’s
circle and apply trigonometry to calculate the angles.
7. Identify orientation of planes for maximum shear stress
(bottom of circle) relative to x-face plane.
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
31
2 S  90deg  2 P  90  33.7  56.3deg
 S  28.15deg CW from the x-face
OR, plane of max shear stress (bottom of circle) is 45 CW in the
real world from the principal stress plane with  P1.
Note:  P   S  45 deg (always !)
Note that the plane of max shear stress will generally have nonzero normal stress! In this case, the plane of max shear stress has
a normal stress  n  35ksi .
8. Draw three free bodies: with the x-y stresses, with the principal
stresses, and with the max shear stresses.
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
20 ksi
The dark face is
the original x-face
as it is rotated.
10 ksi
50 ksi
50 ksi
10 ksi
20 ksi
16.97 ksi
53.03 ksi
16.85
principal stress
planes
53.03 ksi
16.97 ksi
35 ksi
35 ksi
45
18.03 ksi
18.03 ksi
35 ksi
maximum shear
stress planes
35 ksi
28.15
The shear stress
on the rotated xface is negative on
Mohr’s circle and
thus positive on
the stress cube.
32
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
33
Note the following:
 To obtain the orientation for principal stresses, the x-face has
been rotated 16.85 deg CCW in the real world (33.7 deg on
Mohr’s circle).
 To obtain the orientation for maximum shear stresses, the xface has been rotated 28.15 deg CW in the real world (56.3
deg on Mohr’s circle), OR, equivalently a rotation of 45 deg
CW in the real world from the principal stress plane.
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
34
Generalized Plane Stress
In the previous development of the Mohr’s Circle, we
considered only the stresses in the x-y plane that resulted in
two principal stresses  P1 and  P 2 (remember, that
principal stresses occur on planes where there is no shear
stress). We can also use Mohr’s Circle for the case when the
third plane (for example, the z plane) contains only a normal
stress  zz . The stress tensor is
0 
 xx  xy


    yx  yy 0 


0  zz 
 0
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
35
Since there are no shear stresses on the z plane,  zz is a
principal stress, i.e.,  P3   zz .
We note that in the x-y plane, Mohr’s circle is defined by a
circle with coordinates of ( P1 , 0) and ( P 2 , 0). Thus, the
circle is (or can be) defined by the principal stresses. It
follows that the Mohr’s circle for the x-z plane would also
be formed the principal stresses in the x-z plane, in this case
 P1 and  P 3 . Similarly, Mohr’s circle in the y-z plane is
defined by  P 2 and  P 3 . We can draw all three of these
circles on one diagram as shown below.
 2001, W. E. Haisler
36
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
 x ' y ' ( s )
S
( generalized plane stress )
max
( xx ,  xy )
r
 P3
( zz , 0)
 P2
2 P
2 S
( yy ,  xy )
S
max
 P1  x ' x ' ( n )
 2001, W. E. Haisler
37
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
Mohr's Circle Generalized Plane Stress Example. Consider the
plane stress state given by the stress tensor:
20 ksi
10
10
50 10 0 




10
20
0
ksi
  

 0 0 10 
10
50
50 ksi
10
10
y
10
x
20
z
Steps 1-7 will be identical to the previous example for the x-y
plane (since both examples have exactly the same stress
components in the x-y plane). Hence, for the x-y plane we obtain
Mohr’s circle:
 2001, W. E. Haisler
 x' y ' ( s )
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle

Smax
 18.03ksi
50 10 0 
   10 20 0  ksi
 0 0 10 
(35,18.03)
(20,10)
y - face
 x' x' ( n )
(35, 0)

P2
 16.97ksi
38
16.85
2 S

P1
 53.03ksi
(50, -10)
x - face
(35, 18.03)

Smax
 18.03ksi
To complete the solution we note that there are no shear stresses
in the z plane and hence  zz is automatically a principal stress.
Thus,  P 3   zz  10 ksi. Adding the third principal stress to
Mohr’s circle, we obtain:
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle

Smax
 x' y ' ( s )
39
50 10 0 
  10 20 0  ksi
 0 0 10 
 31.52ksi
(35,18.03)
(20,10)
y - face
 x' x' ( n )
(35, 0)

P3
 10ksi

P2
 16.97ksi
16.85
2 S

P1
 53.03ksi
(50, -10)
x - face
(35, 18.03)

Smax
 31.52ksi
Hence, the principal stresses are  P  10, 16.97, 53.03 ksi .
The maximum shear stress is the radius of the largest Mohr’s
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
40
53.03  10
 31.52ksi when generalized
circle. Thus,  S 
max
2
plane stress is considered.
 2001, W. E. Haisler
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
41
Some practice problems for Plane and Generalized Plane Stress:
1.
2.
3.
4.
5.
6.
7.
 xx  50ksi,  yy  10ksi,  xy  20ksi (plane stress in x-y plane)
 yy  50ksi,  zz  10ksi,  yz  20ksi (plane stress in y-z plane)
 xx  10MPa,  yy  50MPa,  xy  20MPa
 xx  10MPa,  yy  50MPa,  xy  20MPa
 xx  10ksi,  yy  50ksi,  xy  0ksi
 xx  0ksi,  yy  0ksi,  xy  20ksi
 xx  50MPa,  yy  50MPa,  xy  0MPa (hydrostatic
compressive stress)
8.  xx  50ksi,  yy  10ksi,  xy  20ksi,  zz  0ksi
(generalized plane stress)
9.  xx  50ksi,  yy  10ksi,  xy  20ksi,  zz  20ksi
(generalized plane stress)
10.  xx  20ksi,  yy  50ksi,  zz  10ksi,  yz  20ksi
 2001, W. E. Haisler
42
Chapter 5: Transformation of Stresses, Principal Stresses and Mohr’s Circle
Note on Three-Dimensional Stress Analysis
For a general stress tensor (3-D), it can be shown that the
principal stresses are defined by the following eigenvalue
problem:
    p  I   0 or
 xx   p
 xy
 xz
 yx
 yy   p
 yz
 zx
 zy
 zz   p
0
The above yields a cubic equation in  p which can be solved
for the three principal stress components. (See Sec. 2.2.7 of
Introduction to Aerospace Structural Analysis by Allen and Haisler).